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znimon
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Solve the nonlinear inequality. Express the solution in interval notation. x^4 is greater than x^2
 one year ago
 one year ago
znimon Group Title
Solve the nonlinear inequality. Express the solution in interval notation. x^4 is greater than x^2
 one year ago
 one year ago

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znimon Group TitleBest ResponseYou've already chosen the best response.0
\[x^4>x^2\]
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
I take the square root of both sides and get the plus or minus square root of x^2 is greater than x, then I take the square root of x^2 and x to get x is greater than the plus or minus square root of x
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
\[x ^{4}>x ^{2}\] \[x ^{4}x ^{2}>0\] \[x ^{2}(x ^{2}1)>0\] \[x ^{2}>0 or x ^{2}>1\] x>0 or x<1 or x>1 then x<1 and x>0 will be the interval...
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
\(x^2(x^21)>0\) divide it by \(x^2\): \((x1)(x+1)>0\) \(x\in(\infty;1)\bigcup(1;\infty)\)
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
why is it greater than 0
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
oh subtract the x^2?
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
yep//....
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
@znimon understood the process?
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
almost, what about after x^2(x^21) is greater than 0
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
\(x^2\) is always bigger than 0. It equals 0 when \(x=0\). But it doesn't satisfy the inequality.
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
took x^2 common in both the terms....
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
yep got that
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
nops.. while splitting the inequality x has to be >0
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
will you break it down after the x^2(x^21) is greater than 0
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
yep.... both the split terms are compared to the inequalities...
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
I understand how it works so I need to find all the solutions for x that satisfy the inequality
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
thanks now I'll try some on my own. Will you recommend a problem that has the same properties?
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
@sriramkumar
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
try this one... \[x ^{8}x ^{2}<0\] remeber x^3 has different properties when compared with x^2 .... have a good time!!!!1
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
@znimon you're welcome :)))))))))))))))))
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
\[\ \ x^4\ge x^2\\ \ \ x^2[x^2]\ge x^2\\x^2\text{ is either 0 nonzero.}\\ \text{if }x^2\text{ is 0,}\\ \ \ 0\times0=0\\ \ \ 0 = 0\\ \text{therefore }x^2[x^2]\ge x^2\\ \text{otherwise,}\\ \ \ x^2\ge1\\ \ \ x\ge\sqrt1\]
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
is either 0 or nonzero **
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
solution is \[x <0, x <1] so (negative infinity, to 0 not including zero)?
 one year ago

znimon Group TitleBest ResponseYou've already chosen the best response.0
for x^3x^2 is less than 0
 one year ago
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